I eventually figured it out by solving it for the case of 1 blue-eyed person, 2 blue-eyed persons, 3 blue-eyed persons and 4 blue-eyed persons, until I caught onto the logic.
It also helped me to paraphrase the Guru’s comment as “At least one blue-eyed person can go home now” or something like that.
The rest of the Brown-eyed people do not leave because they can’t be sure what their eye colour is.
Welcome to the world of logic puzzles. You should assume unless otherwise advised that all citizens have perfect powers of deduction, and tell the truth, or not, in a perfectly describable fashion. If the citizens in question are pirates, then you must assume that they are incredibly bloodthirsty, greedy, selfish, and malicious. The king of Puzzlania routinely executes people for failing to guess what colour hat they’re wearing. If you’re a prisoner in Puzzlania, forget it. You’re going to have to work out some ludicrously difficult brainteaser if you’re ever going to see daylight. And so on.
It’s probably not a very nice place to live, but it makes for some interesting puzzles.
(anyone know of a good book with puzzles along this vein that a kid might enjoy?)
What is the Name of This Book? by Raymond Smullyan.
Collection of logic puzzles (mostly of the truth-tellers-and-liars variety) all done up into entertaining stories. In one chapter, for example, he visits Transylvania to investigate for himself whether Dracula was really ever destroyed or not – only to find that all Transylvanians are:
– a. Knights (truth-tellers) or knaves (liars), all of whom tell the truth or lies according to the truth as they know it.
– b. Sane (who know what the truth is) or insane (who are completely deluded and believe that all true propositions are false, and all false propositions are true).
Thus, all insane knights always say what is false (because that’s what they believe is true), while all insane knaves say what is true (because they believe the opposite, but lie according to what they believe).
The book also has chapters that go into deeper logic subjects, in particular Godelian islands and such. But it’s all fun reading.
ETA: Also contains a chapter of silly (mostly well-known) riddles, and a chapter of funny anecdotes about famous mathematicians, physicists, logicians, etc.
This is correct: If there are four blue-eyed people, then nobody is surprised by anything that happens on Night 2, because everyone already knew that nobody would leave on Night 2. But on Night 3, the blue-eyed folks don’t know in advance what’s going to happen, and so learn something new (namely, that they should leave), and on Night 4, the non-blues don’t know in advance what’s going to happen, and so learn something new (namely, that they should stay).
I am.
I’m not following who talks to whom to verify what.
I had no trouble with the blue eyed islanders.
As an aside, Indistinguishable, are you familiar with Adinkra diagrams? They seem to have a structure somewhat similar to your islands-and-bridges problem, and are used in some String Model calculations. Our colloquium speaker was talking (in part) about them today, and mentioned that they’re only just starting to come to the attention of mathematicians.
Recommendation heartily seconded. He did a series of such puzzle books and they’re all great fun. It’s been far too long since I visited the Island of the Knights and Knaves.
Another contender BTW for The Hardest Logic Puzzle Ever.
Another of Smullyan’s logic puzzle books is To Mock A Mockingbird.
Here’s a similar puzzle that I devised earlier today in order to mimic the eye-color puzzle.
Suppose you were a contestant on a game show with twenty teammates. Your team will be asked twenty factual questions and for each question, one and only one teammate will write down an answer on a sheet of paper. For the next question, they’ll pass it to their teammate who will write their answer to the new question below the previous answer. Simple, right?
Problem is, your opponents get to pick who will answer each question. And since the captain of your opponent’s team is the sexy young red-head you unwittingly bedded last night, you suddenly break into a cold sweat as you recall telling her what categories of knowledge each of your teammates specialize in. There are twenty categories of questions, but you thought it wise for each teammate to memorize the answers in only 19 categories, each. So every person has a gaping hole in their knowledge bank for only one category. And that’s the person she’ll choose to answer the question for that category.
So you’re in the strange situation that everyone knows the answer to every question except their own. For their own question, they can only guess with a 50/50 shot of guessing correctly.
Now the question: Is it possible that by the time you get the paper on the 20th question that your team will have answered incorrectly for every preceding answer? And if so, is it possible for you to answer incorrectly as well? So is it then possible that your team will score absolutely zero points?
Why all the confusion when **Chronos’ **mathematical proof is in front of you? The only thing I would change in the above proof is the Increment step. As written it’s not obvious exactly where or how the induction hypothesis is applied. I would word the proof as follows (with a nod to Chronos):
Proposition P(N): “For any whole number N >= 1, if there are N blue-eyed islanders, then all blue-eyed islanders will leave on day N.”
Prove P(1): If N=1, then the single blue-eye will know that he must be the blue-eye person referred to by the Guru, and leaves on night 1. Therefore, P(1) is true.
Prove P(N) implies P(N+1): Assume that P(N) is true (this is the induction hypothesis). We must now prove that P(N+1) is true. Assume that there are N+1 blue-eyes. Each blue-eye would see N other blue-eyes. Each blue-eye would apply P(N) to infer that those other N blue-eyes would have all left the previous day if they were the totality of blue-eyes. Since they did not leave each blue-eye will conclude that he is also a blue-eye. And thus, each of the N+1 blue-eyes will declare that he’s a blue-eye and leave the island. (Note that a non-blue-eye cannot use this logic.) Therefore, P(N+1) is true. And the induction proof is complete.
The induction explanation is fine for showing *how *the islanders realise that they have blue eyes, but I don’t think it’s so good at explaining *why *- the question of what new information, if any, the Guru brings to the situation, which is what the subsequent discussion has mostly been about.
Ultimately, the problem is that you (a blue) can count 99, but you don’t know if the blues you see count 99 or only 98. They can’t communicate their count to you. The new info the Guru adds is a start time, and the color to count. The Guru is starting the count at 1. “I see 1 blue eyed person.” From that point, every one else can count blue eyes and days, and see if the blues have left yet.
Until some event can start the count, the chain of logic it’s based on can’t be used.
I’m not going to attempt any explanation, but I will point to the bit which helped me understand what was going on, and especially what extra you get from the Guru.
Emphasis mine.
i’m still stuck here but it’s my first post. I have read the entire thread most of it OVER AND OVER. what if the one person the guru is talking about is the same 1 person you are looking at with blue eyes does the answer still play out? still possible?.
It doesn’t matter who the person with blue eyes is. That makes no difference to the information given.
Maybe I’m missing the point here, but it seems obvious that is possible, if perhaps unlikely, that everybody will guess wrong.
I still (yes still so sorry for my ignorance) only see 2 outcomes. I see
You are the only blue eyed person and since you see no other blue eyes you know the guru is talking about you.
there is at least one other blue eyed person you see which the guru may be talking about which in that case you know the same exact thing he knows and just said being: he see’s at least one blue eyed person.
2.5 what he said means nothing because you already knew that he knew that even before he said it unless he’s blind there’s no way he didn’t already see one person. and since you already knew what he was looking at because you are looking at at LEAST that one person also still suck in the same position? day’s won’t help?.
that’s my conclusion so far. (i’m still trying. but you simplest breakdown still too complicated for me apparently).
In that case, you must consider the possibility that the other person is the only blue eyed person, and therefore does not see anyone with blue eyes.